How do we find localization of a ring? Let $R = \mathbb Z/6 \mathbb Z$  and let $S=\{1,2,4\}$. Show $ S^{-1}R \cong \mathbb Z/3 \mathbb Z $.
I understand that in $S^{-1}R$ the elements of $S$ must be invertible but I'm not getting a way to prove above isomorphism. Any ideas? 
 A: We want to find $(\mathbb{Z}/6\mathbb{Z})_2$. Now $$(\mathbb{Z}/6\mathbb{Z})_2\cong(\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})_2\cong (\mathbb{Z}/3\mathbb{Z})_2\times(\mathbb{Z}/2\mathbb{Z})_2\cong \mathbb{Z}/3\mathbb{Z}.$$ In the last step $(\mathbb{Z}/2\mathbb{Z})_2\cong 0$ because we are inverting $0$ and $(\mathbb{Z}/3\mathbb{Z})_2\cong \mathbb{Z}/3\mathbb{Z}$ because we are inverting something which is already invertible.

Answer to comment: take $M,N$ two $A$-modules, $S\subseteq A$ multiplicative. Now $$S^{-1}(M\oplus N)\cong (M\oplus N)\otimes S^{-1}A\cong (M\otimes S^{-1}A)\oplus (N\otimes S^{-1}A)\cong S^{-1}M\oplus S^{-1}N.$$ Notice that $\oplus$ and $\times$ are the same since we are dealing with a finite number of modules. 
If you don't know about tensor products yet, just show that the natural map $$S^{-1}(M\oplus N)\to S^{-1}M\oplus S^{-1}N$$ given by $\frac{(m,n)}{s} \mapsto (\frac{m}{s},\frac{n}{s})$ is a well-defined isomorphism of $A$-modules.
A: We have that $\frac 3 1= \frac 6 2=\frac 0 1$ and $\frac 1 1\neq \frac 0 1$. 
Since $1$ generates the original ring as a group,$ \frac 1 1$ generates the localization as a group, so it only has 3 elements.
There is only one ring with three elements: $\Bbb Z/3 \Bbb Z$.
